# Integration calculator error? doubt it.

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• Jan 27th 2009, 05:33 PM
pberardi
Integration calculator error? doubt it.
Hi,
When integrating 2sin(x)cos(x), I used a simple u sub and got sin^2(x), no big deal. However when I pop this into wolfram, it results in (-1/2)cos(2x). This is the link Wolfram Mathematica Online Integrator
I take the derivative of what wolfram gives me and it certainly is not 2sin(x)cos(x). Can someone tell me what I did wrong please?
• Jan 27th 2009, 05:39 PM
skeeter
you didn't do anything wrong.

graph $y = \sin^2{x}$ and $y = -\frac{\cos(2x)}{2}$

... what do you see?
• Jan 27th 2009, 05:40 PM
Chris L T521
Quote:

Originally Posted by pberardi
Hi,
When integrating 2sin(x)cos(x), I used a simple u sub and got sin^2(x), no big deal. However when I pop this into wolfram, it results in (-1/2)cos(2x). This is the link Wolfram Mathematica Online Integrator
I take the derivative of what wolfram gives me and it certainly is not 2sin(x)cos(x). Can someone tell me what I did wrong please?

Both are correct.

Differentiate your solution, and you get $2\sin x\cos x$

However, when you differentiate $-\tfrac{1}{2}\cos(2x)$, you get $\sin(2x)$

But from trig, $\sin(2x)=2\sin x\cos x$.

This shows that both solutions are the same :)

Does this clarify things?

EDIT: Skeeter beat me....
• Jan 27th 2009, 05:54 PM
Soroban
Hello, pberardi!

Quote:

When integrating $2\sin x\cos x$, I used a simple u sub and got $\sin^2\!x$

However when I pop this into wolfram, it results in: . $-\tfrac{1}{2}\cos(2x)$

This is a classic problem/puzzle . . .

You had: . $2\int\sin x(\cos x\,dx)$

. . You let: $u \,=\,\sin x\quad\Rightarrow\quad du \,=\,\cos x\,dx$

. . Substitute: . $2\int u\,du \:=\:u^2+C$

. . Back-substitute: . $\boxed{\sin^2\!x + C}$

We also have: . $2\int \cos x(\sin x\,dx)$

. . Let: $u \,=\,\cos x\quad\Rightarrow\quad du \,=\,-\sin x\,dx \quad\Rightarrow\quad \sin x\,dx \,=\,-du$

. . Substitute: . $2\int u(-du) \:=\:-\int u\,du \:=\:-u^2 + C$

. . Back-substitute: . $\boxed{-\cos^2\!x + C}$

We have the identity: . $2\sin\theta\cos\theta \:=\:\sin2\theta$

So: . $\int2\sin x\cos x\,dx \:=\:\int\sin2x\,dx \:=\:\boxed{-\tfrac{1}{2}\cos2x + C}$

And all three answers are correct!

• Jan 27th 2009, 06:09 PM
pberardi
Ok that all makes sense, so they are the same function. But why is it that when I plug in pi/4 in sin^2(x) I get 1/2 and when I plug in pi/4 into (-1/2)cos(2x), I get 0? Shouldn't they give me the same answer? I know this is silly and I am probably just making a stupid error somewhere but I dont know where.
• Jan 28th 2009, 01:49 AM
mr fantastic
Quote:

Originally Posted by pberardi
Ok that all makes sense, so they are the same function. But why is it that when I plug in pi/4 in sin^2(x) I get 1/2 and when I plug in pi/4 into (-1/2)cos(2x), I get 0? Shouldn't they give me the same answer? I know this is silly and I am probably just making a stupid error somewhere but I dont know where.

If you're using the anti-derivative to evaluate a definite integral, then you will be substituting two numbers (the upper terminal and the lower terminal). Then you will get the same answer, assuming you perform the calculation correctly.
• Jan 28th 2009, 04:21 AM
skeeter
Quote:

Originally Posted by pberardi
Ok that all makes sense, so they are the same function. But why is it that when I plug in pi/4 in sin^2(x) I get 1/2 and when I plug in pi/4 into (-1/2)cos(2x), I get 0? Shouldn't they give me the same answer? I know this is silly and I am probably just making a stupid error somewhere but I dont know where.

they are not the same function ... did you look at their graphs as I recommended?

graph them. what do you see?